Solution:

Since $3^2 + 4^2 = 5^2,$ angle $B$ is right, and as $\angle EBD = 90^\circ$ is inscribed in $\omega,$ the chord $ED$ is a diameter. Hence $\angle EFD = \angle EGD = 90^\circ.$ From $EF = DF,$ triangle $EFD$ is an isosceles right triangle, so $EF = DF = \frac{DE}{\sqrt{2}}$ and $\angle FED = 45^\circ;$ from $DG : EG = 3 : 4$ and $DG^2 + EG^2 = DE^2$ we get $DG = \frac{3}{5}DE$ and $EG = \frac{4}{5}DE.$ On $\omega$ the order is $E, F, G, D$ (both $F$ and $G$ lie on the arc opposite $B,$ and $\angle FED = 45^\circ$ exceeds $\angle GED = \arcsin\frac{3}{5},$ so $F$ is farther from $D$).

Line $AC$ is the line $FG,$ so the distances from $E$ and $D$ to it follow from the angles of cyclic quadrilateral $EFGD.$ At $F:$ $\angle EFG = 180^\circ - \angle GDE$ and $\sin\angle GDE = \frac{EG}{DE} = \frac{4}{5},$ so the distance from $E$ is $EF \sin\angle EFG = \frac{DE}{\sqrt{2}} \cdot \frac{4}{5} = \frac{2\sqrt{2}}{5}DE.$ At $G:$ $\angle FGD = 180^\circ - \angle FED = 135^\circ,$ so the distance from $D$ is $DG \sin\angle FGD = \frac{3}{5}DE \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{10}DE.$

Now place $B = (0,0),$ $C = (4,0),$ $A = (0,3),$ so $E = (0, e),$ $D = (d, 0),$ line $AC:$ $3x + 4y = 12,$ and $DE^2 = d^2 + e^2.$ Setting $k = \frac{\sqrt{2}}{2}DE,$ the two distance formulas read $\frac{12 - 4e}{5} = \frac{2\sqrt{2}}{5}DE$ and $\frac{12 - 3d}{5} = \frac{3\sqrt{2}}{10}DE,$ which give $e = 3 - k$ and $d = 4 - k.$ Then $DE^2 = 2k^2$ becomes $2k^2 = (3-k)^2 + (4-k)^2 = 2k^2 - 14k + 25,$ so $k = \frac{25}{14}$ and $DE = \sqrt{2}\,k = \frac{25\sqrt{2}}{14}.$ Therefore $a + b + c = 25 + 2 + 14 = 41.$